A force of 1N was applied to a rolling ball with a mass of 1 kg, under the action of which the ball stopped

A force of 1N was applied to a rolling ball with a mass of 1 kg, under the action of which the ball stopped, having traveled a path of 1 m. Determine the speed with which the ball was moving before the start of deceleration.

1) By condition, only a force stopping it with a magnitude of F = 1 N acts on the ball, so we write Newton’s second law in an impulse formulation: – F = (p – p0) / t, – F = (m * vk. – m * v0) / t, since the final speed vk. = 0 m / s, then: – F = (0 – m * v0) / t, – F * t = – m * v0, m * v0 = F * t (t is the time it took for the ball to stop completely ).
2) Let us write down the kinematic equation describing the stop of the ball: x = v0 * t – a * t2 / 2.
We express from the final equation of item 1 (m * v0 = F * t) the variable t: t = m * v0 / F.
Also, according to Newton’s second law, F = m * a, a = F / m. Substituting this into the kinematic equation, we have:
x = (m * v0 ^ 2) / (2F), whence v0 = √ ((2F * x) / m) = √ ((2 * 1 N * 1 m) / 1 kg) = √ (2) m / c.
ANSWER: v0 = √ (2) m / s.